Simplify; express your answer in exponential form. Assume $k\neq 0, a\neq 0$. $\dfrac{{(k^{-5})^{-3}}}{{ka^{-2}}}$
Answer: To start, try working on the numerator and the denominator independently. In the numerator, we have ${k^{-5}}$ to the exponent ${-3}$ . Now ${-5 \times -3 = 15}$ , so ${(k^{-5})^{-3} = k^{15}}$ In the denominator, we can use the distributive property of exponents. ${ka^{-2} = ka^{-2}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(k^{-5})^{-3}}}{{ka^{-2}}} = \dfrac{{k^{15}}}{{ka^{-2}}}$ Break up the equation by variable and simplify. $\dfrac{{k^{15}}}{{ka^{-2}}} = \dfrac{{k^{15}}}{{k}} \cdot \dfrac{{1}}{{a^{-2}}} = k^{{15} - {1}} \cdot a^{- {(-2)}} = k^{14}a^{2}$.